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If $\tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3$, then which of the following is equal to 1 ?
Options:
Solution:
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Verified Answer
The correct answer is:
$\tan 3 \theta$
Given,
$$
\begin{aligned}
& \tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3 \\
& \Rightarrow \tan \theta+\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}+\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}=3 \\
& \Rightarrow \quad \tan \theta+\frac{8 \tan \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad \frac{9 \tan \theta-3 \tan ^3 \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad 3 \tan 3 \theta=3 \Rightarrow \tan 3 \theta=1 \\
&
\end{aligned}
$$
Hence, option (2) is correct.
$$
\begin{aligned}
& \tan \theta+\tan \left(\theta+\frac{\pi}{3}\right)+\tan \left(\theta+\frac{2 \pi}{3}\right)=3 \\
& \Rightarrow \tan \theta+\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}+\frac{\tan \theta-\sqrt{3}}{1+\sqrt{3} \tan \theta}=3 \\
& \Rightarrow \quad \tan \theta+\frac{8 \tan \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad \frac{9 \tan \theta-3 \tan ^3 \theta}{1-3 \tan ^2 \theta}=3 \\
& \Rightarrow \quad 3 \tan 3 \theta=3 \Rightarrow \tan 3 \theta=1 \\
&
\end{aligned}
$$
Hence, option (2) is correct.
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